On Janowski functions associated with (n,m)-symmetrical functions

Abstract

The aim in the present work is to introduce and study new subclasses of analytic functions that are defined by using the generalized classes of Janowski functions combined with the $(n,m)$-symmetrical functions, that generalize many others defined by different authors. We gave a representation theorem for these classes, certain inherently properties, while covering and distortion properties are also pointed out.

Keywords:

• Janowski functions
• differential subordination
• starlike functions
• convex functions
• m-fold symmetric domain
• (n,m)-symmetrical functions

2010 Mathematics Subject Classification:

• Primary 30C45

1. Introduction

Let $\mathcal{H}\left(\mathrm{U}\right)$ be the space of all analytic functions in the open unit disk $\mathrm{U}:=\{z\in \mathbb{C}:|z|<1\},$ and let $\mathcal{A}$ denote the class of functions $f\in \mathcal{H}\left(\mathrm{U}\right)$ normalized with $f\left(0\right)=f^{\prime }\left(0\right)-1=0$. Denote by Ω for the class of Schwarz functions, that is (1) $\mathrm{\Omega }:=\{w\in \mathcal{A},w(0)=0,|w(z)|<1,z\in \mathrm{U}\}.$(1) For f and g are two analytic functions in $\mathrm{U}$, we say that the function f is subordinate to the function g in $\mathrm{U}$, if there exists a function $w\in \mathrm{\Omega }$, such that $f\left(z\right)=g\left(w\right(z\left)\right)$ for all $z\in \mathrm{U}$, and we denote this by $f\left(z\right)\prec g\left(z\right)$. Furthermore, if g is univalent in $\mathrm{U}$, then the subordination is equivalent to $f\left(0\right)=g\left(0\right)$ and $f\left(\mathrm{U}\right)\subset g\left(\mathrm{U}\right)$. (see, for details, [1])

Using the notation of the subordination, let define the class $\mathcal{P}$ of functions with positive real parts (or Carathéodory functions) in $\mathrm{U}$:

Definition 1.1

[2]

Let $\mathcal{P}$ denote the class of functions $p\in \mathcal{H}\left(\mathrm{U}\right)$ satisfying $p\left(0\right)=1$ and $\mathrm{Re}p\left(z\right)>0$ for all $z\in \mathrm{U}$.

From the above-mentioned reasons, it follows that any function $p\in \mathcal{P}$ has the representation $p\left(z\right)=\left(1+w\left(z\right)\right)/\left(1-w\left(z\right)\right)$, for some $w\in \mathrm{\Omega }$.

The class of functions with positive real part plays a significant role in complex function theory. Its significance can be seen from the fact that all simple subclasses of the class of univalent functions have been defined by using the concept of the class of functions with positive real part like the classes ${\mathcal{S}}^{\ast }$, $\mathcal{C}$, $S^k$ which are respectively the class of starlike, convex functions and the class of starlike functions with respect to symmetric points, etc., have been defined by using the class $\in \mathcal{P}$.

Definition 1.2

1. Like in [3], let $\mathcal{P}[X,Y]$, with $-1\le Y<X\le 1$, denote the class functions $p\in \mathcal{H}\left(\mathrm{U}\right)$ that satisfy the subordination condition $p\left(z\right)\prec \left(1+Xz\right)/\left(1+Yz\right)$.

2. The class of generalized Janowski type functions $\mathcal{P}[X,Y,\gamma ]$ was introduced in [4] as follows: for arbitrary fixed numbers $X,Y,\gamma$, with $-1\le Y<X\le 1$, $0\le \gamma <1$, $p\in \mathcal{P}[X,Y,\gamma ]⟺p\left(z\right)\prec \frac{1+\left[\right(1-\gamma )X+\gamma Y]z}{1+Yz}.$

3. Like in [4], a function $f\in \mathcal{A}$, then $f\in {\mathcal{S}}^{\ast }[X,Y,\gamma ]⟺\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\in \mathcal{P}[X,Y,\gamma ].$

Definition 1.3

1. For the positive integer number m, a domain $\mathbf{D}\subset \mathbb{C}$ is said to be m-fold symmetric domain, if a rotation of $\mathbf{D}$ about the origin through and angle $2\pi /m$ carries $\mathbf{D}$ onto itself.

2. A function $f:\mathbf{D}\to \mathbb{C}$, where $\mathbf{D}$ is a m-fold symmetric domain, is said to be m-fold symmetrical function if, $f\left(e^{2\pi i/m}z\right)=e^{2\pi i/m}f\left(z\right),z\in \mathbf{D}.$ The set of all m-fold symmetrical denoted by ${\mathcal{S}}^m$,

The theory of $(n,m)$-symmetrical functions for $(n=0,1,\dots ,m-1)$ and ($m=1,2,\dots$) is a generalization of the notion of odd, even, m-symmetrical functions.

Definition 1.4

Let $\epsilon =e^{2\pi i/m},n=0,1,\dots ,m-1$, where $m\in \mathbb{N}$ and $m\ge 2$.

A function $f:\mathbf{D}\to \mathbb{C}$, where $\mathbf{D}$ is a m-fold symmetric domain, is called $(n,m)$-symmetrical function if $f\left(\epsilon z\right)={\epsilon }^{n}f\left(z\right),z\in \mathbf{D}$.

Denoted be ${\mathcal{S}}^{(n,m)}$ for the set of all $(n,m)$-symmetrical functions. Let us observe that the sets ${\mathcal{S}}^{(0,2)}$, ${\mathcal{S}}^{(1,2)}$, and ${\mathcal{S}}^{(1,m)}$ are well-known families of odd functions, of even functions and of $m-$symmetrical functions, respectively.

The following decomposition theorem holds:

Theorem 1.1

[5]

Let $\mathbf{D}$ be a m-fold symmetric domain, the for every function $f:\mathbf{D}\to \mathbb{C}$, can be written in the form $f\left(z\right)=\sum _{n=0}^{m-1}{f}_{n,m}\left(z\right),z\in \mathbf{D},$ and this partition is unique sequence of $(n,m)$-symmetrical functions. (2) $\begin{array}{rl}& f_{n,m}\left(z\right)=\frac{1}{m}\sum _{\nu =0}^{m-1}{\epsilon }^{-\nu n}f\left({\epsilon }^{\nu }z\right),z\in \mathbf{D},\\ & (m=2,3,\dots ;n=0,1,2,\dots ,m-1).\end{array}$(2)

Al Sarari and Latha [6] introduced the classes ${\mathcal{S}}^{(n,m)}[X,Y]$ and ${\mathcal{K}}^{(n,m)}[X,Y]$ which are the classes of Janowski type functions with respect to $(n,m)$-symmetric points.

By using the theory of $(n,m)$-symmetrical functions and the generalized Janowski type functions, we will define the following class:

Definition 1.5

A function $f\in \mathcal{A}$ is said to belongs to the class ${\mathcal{S}}^{(n,m)}[X,Y,\gamma ]$, with $-1\le Y<X\le 1$ and $0\le \gamma <1$, if $\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\prec \frac{1+\left[\right(1-\gamma )X+\gamma Y]z}{1+Yz},$ where the function $f_{n,m}\left(z\right)$ is defined by (2(2) $\begin{array}{rl}& f_{n,m}\left(z\right)=\frac{1}{m}\sum _{\nu =0}^{m-1}{\epsilon }^{-\nu n}f\left({\epsilon }^{\nu }z\right),z\in \mathbf{D},\\ & (m=2,3,\dots ;n=0,1,2,\dots ,m-1).\end{array}$(2) ).

Remark 1.1

By applying the definition of the subordination we can easily obtain that the equivalent condition for a function f belonging to the class ${\mathcal{S}}^{(n,m)}[X,Y,\gamma ]$, with $-1\le Y<X\le 1$ and $0\le \gamma <1$, is $\left|\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}-1\right|<\left|\left[\right(1-\gamma )X+\gamma Y]-Y\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\right|,z\in \mathrm{U}.$

Note that special values of n, m, γ, X and Y yield the following classes that have been previously introduced by different authors:

1. ${\mathcal{S}}^{(n,m)}[X,Y,0]=:{\mathcal{S}}^{(n,m)}[X,Y]$ is the class studied by Al Sarari and Latha [6].

2. ${\mathcal{S}}^{(1,m)}[1,-1,0]={\mathcal{S}}_m^{\ast }(1,-1)=:{\mathcal{S}}_m^{\ast }$ is the class is introduced by Sakaguchi [7].

3. ${\mathcal{S}}^{(1,m)}[X,Y,0]=:{\mathcal{S}}^{\left(m\right)}[X,Y]$ is the class was introduced by Ohang and Youngjae in [8].

The following lemmas are important to proof our results:

Lemma 1.1

[4, Corollary 1]

For a function $f\in {\mathcal{S}}^{\ast }[X,Y,\gamma ],$ then $f\left(z\right)=\left\{\begin{array}{ll}z\exp \left[(1-\gamma )Xw\left(z\right)\right],& \mathrm{i}\mathrm{f}Y=0,\\ z{\left(1+Yw\left(z\right)\right)}^{\left((1-\gamma )(X-Y)\right)/Y},& \mathrm{i}\mathrm{f}Y\ne 0,\end{array}\right.$ for some $w\in \mathrm{\Omega },$ where Ω was defined by (1(1) $\mathrm{\Omega }:=\{w\in \mathcal{A},w(0)=0,|w(z)|<1,z\in \mathrm{U}\}.$(1) ).

Lemma 1.2

[9, Lemma 2.3]

For $p\in \mathcal{P}[X,Y,\gamma ],$ then $\begin{array}{rl}& \frac{1-\left[\right(1-\gamma )X+\gamma Y]r}{1-Yr}\le |p\left(z\right)|\\ & \le \frac{1+\left[\right(1-\gamma )X+\gamma Y]r}{1+Yr},|z|\le r<1.\end{array}$

2. Main results

Theorem 2.1

If $f\in {\mathcal{S}}^{(n,m)}[X,Y,\gamma ]$, then (3) $f_{n,m}\left(z\right)=\left\{\begin{array}{ll}z\exp \left[\right(1-\gamma \left)Yw\right(z\left)\right],& \mathrm{i}\mathrm{f}Y=0,\\ z{\left(1+Yw\left(z\right)\right)}^{\left((1-\gamma )(X-Y)\right)/Y},& \mathrm{i}\mathrm{f}Y\ne 0,\end{array}\right.$(3) where $f_{n,m}\left(z\right)$ is defined by (2(2) $\begin{array}{rl}& f_{n,m}\left(z\right)=\frac{1}{m}\sum _{\nu =0}^{m-1}{\epsilon }^{-\nu n}f\left({\epsilon }^{\nu }z\right),z\in \mathbf{D},\\ & (m=2,3,\dots ;n=0,1,2,\dots ,m-1).\end{array}$(2) ), and for some $w\in \mathrm{\Omega }$.

Proof.

Let $f\in {\mathcal{S}}^{(n,m)}[X,Y,\gamma ]$, we can get (4) $\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\prec \frac{1+\left[\right(1-\gamma )X+\gamma Y]z}{1+Yz}.$(4) Replacing z by ${\epsilon }^{\nu }z$ in (4(4) $\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\prec \frac{1+\left[\right(1-\gamma )X+\gamma Y]z}{1+Yz}.$(4) ), $\begin{array}{rl}\frac{{\epsilon }^{\nu }z{f}^{\prime }\left({\epsilon }^{\nu }z\right)}{f_{n,m}\left({\epsilon }^{\nu }z\right)}& \prec \frac{1+\left[\right(1-\gamma )X+\gamma Y]{\epsilon }^{\nu }z}{1+Y{\epsilon }^{\nu }z}\\ & \prec \frac{1+\left[\right(1-\gamma )X+\gamma Y]z}{1+Yz},\end{array}$ hence (5) $\frac{{\epsilon }^{\nu -\nu j}z{f}^{\prime }\left({\epsilon }^{\nu }z\right)}{f_{n,m}\left(z\right)}\prec \frac{1+\left[\right(1-\gamma )X+\gamma Y]z}{1+Yz},$(5) Letting $\nu =0,1,\dots ,m-1$ in (5(5) $\frac{{\epsilon }^{\nu -\nu j}z{f}^{\prime }\left({\epsilon }^{\nu }z\right)}{f_{n,m}\left(z\right)}\prec \frac{1+\left[\right(1-\gamma )X+\gamma Y]z}{1+Yz},$(5) ), and since $\mathcal{P}[X,Y,\gamma ]$ is a convex set, we deduce that $\frac{z{\displaystyle \frac{1}{m}}\sum _{\nu =0}^{m-1}{\epsilon }^{\nu -\nu n}{f}^{\prime }\left({\epsilon }^{\nu }z\right)}{f_{n,m}\left(z\right)}\prec \frac{1+\left[\right(1-\gamma )X+\gamma Y]z}{1+Yz},$ or equivalently $\frac{z{f}_{n,m}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\prec \frac{1+\left[\right(1-\gamma )X+\gamma Y]z}{1+Yz},$ that is $f_{n,m}\in {\mathcal{S}}^{\ast }[X,Y,\gamma ]$, and by Lemma 1.1 we finally obtain our result.

Theorem 2.2

For $f\in {\mathcal{S}}^{(n,m)}[X,Y,\gamma ],$ with $-1\le Y<X\le 1$ and $0\le \gamma <1$. Then, $f\left(z\right)=\left\{\begin{array}{ll}\begin{array}{c}{\int }_0^z[1+X(1-\gamma \left)\tilde{w}\right(\zeta \left)\right]\\ \exp \left[\right(1-\gamma \left)Xw\right(\zeta \left)\right]\mathrm{d}\zeta \end{array},& \mathrm{i}\mathrm{f}Y=0,\\ \begin{array}{c}{\int }_0^z\frac{1+\left[\right(1-\gamma )X+\gamma Y]\tilde{w}\left(\zeta \right)}{1+Y\tilde{w}\left(\zeta \right)}\\ {\left(1+Yw\left(\zeta \right)\right)}^{\left((1-\gamma )(X-Y)\right)/Y}\mathrm{d}\zeta \end{array},& \mathrm{i}\mathrm{f}Y\ne 0.\end{array}\right.$ For some $\tilde{w},w\in \mathrm{\Omega }$.

Proof.

Supposing that $f\in {\mathcal{S}}^{(n,m)}[X,Y,\gamma ]$, then there exists a function $\tilde{w}\in \mathrm{\Omega }$, such that $\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}=\frac{1+\left[\right(1-\gamma )X+\gamma Y]\tilde{w}\left(z\right)}{1+Y\tilde{w}\left(z\right)},z\in \mathrm{U}.$ Combining the above relation with Theorem 2.1, we have $f^{\prime }\left(z\right)=\left\{\begin{array}{ll}\begin{array}{c}[1+X(1-\gamma \left)\tilde{w}\right(z\left)\right]\\ \exp \left[\right(1-\gamma \left)Xw\right(z\left)\right]\end{array},& \mathrm{i}\mathrm{f}Y=0,\\ \begin{array}{c}\frac{1+\left[\right(1-\gamma )X+\gamma Y]\tilde{w}\left(z\right)}{1+Y\tilde{w}\left(z\right)}\\ {\left(1+Yw\left(z\right)\right)}^{\left((1-\gamma )(X-Y)\right)/Y}\end{array},& \mathrm{i}\mathrm{f}Y\ne 0,\end{array}\right.$ and integrating the above relations we obtain our result.

Theorem 2.3

Let $f\in {\mathcal{S}}^{(n,m)}[X,Y,\gamma ]$ and $f_{\mu }\left(z\right):=\mu f\left(z\right)+(1-\mu )z,$ with $0<\mu <1$. Then,

1. $f_{\mu }\in {\mathcal{S}}^{(n,m)}[X,0,\gamma ]$, if Y = 0;

2. $f_{\mu }\in {\mathcal{S}}^{(n,m)}[X,Y,\gamma ]$, for $|z|<1/Y\sin \left(Y/\right((1-\gamma )X+\gamma Y\left)\right(\pi /2\left)\right)$, if Y >0;

3. $f_{\mu }\in {\mathcal{S}}^{(n,m)}[X,Y,\gamma ],$ for $|z|<1/Y\sin \left(Y/\right(2Y-\left[\right(1-\gamma )X+\gamma Y]\left)\right(\pi /2\left)\right)$, if Y <0.

Proof.

Since $f\in {\mathcal{S}}^{(n,m)}[X,Y,\gamma ]$, then $\begin{array}{rl}\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}& \prec \frac{1+\left[\right(1-\gamma )X+\gamma Y]z}{1+Yz},\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\\ f_{n,m,\mu }\left(z\right)& =\frac{1}{m}\sum _{\nu =0}^{m-1}{\epsilon }^{-\nu n}{f}_{\mu }\left({\epsilon }^{\nu }z\right).\end{array}$ Thus, $\begin{array}{rl}{f}_{n,m,\mu }\left(z\right)& =\mu f_{n,m}\left(z\right)+(1-\mu )z,\\ z{f}_{\lambda }^{\prime }\left(z\right)& =(1-\mu )z+\mu z{f}^{\prime }\left(z\right),\end{array}$ hence (6) $\frac{z{f}_{\mu }^{\prime }\left(z\right)}{f_{n,m,\mu }\left(z\right)}=\frac{(1-\mu ){\displaystyle \frac{z}{f_{n,m}\left(z\right)}}+\mu {\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}}}{(1-\mu ){\displaystyle \frac{z}{f_{n,m}\left(z\right)}}+\mu }.$(6)

(i) For Y = 0 it is sufficient to show that (7) $\left|\frac{(1-\mu ){\displaystyle \frac{z}{f_{n,m}\left(z\right)}}+\mu {\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}}}{(1-\mu ){\displaystyle \frac{z}{f_{n,m}\left(z\right)}}+\mu }-1\right|<(1-\gamma )X,z\in \mathrm{U}.$(7) From $f\in {\mathcal{S}}^{(n,m)}[x,Y,\gamma ]$ we have $\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\prec 1+(1-\gamma )Xz$ which implies $\left|\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}-1\right|<(1-\gamma )X,z\in \mathrm{U},$ and according to (3(3) $f_{n,m}\left(z\right)=\left\{\begin{array}{ll}z\exp \left[\right(1-\gamma \left)Yw\right(z\left)\right],& \mathrm{i}\mathrm{f}Y=0,\\ z{\left(1+Yw\left(z\right)\right)}^{\left((1-\gamma )(X-Y)\right)/Y},& \mathrm{i}\mathrm{f}Y\ne 0,\end{array}\right.$(3) ) $\frac{f_{n,m}\left(z\right)}{z}\prec \exp \left[\right(1-\gamma \left)Xz\right],$ hence, there exists a Schwarz function $w\in \mathrm{\Omega }$ such that $\frac{f_{n,m}\left(z\right)}{z}=\exp \left[\right(1-\gamma \left)Xw\right(z\left)\right],z\in \mathrm{U}.$ Thus, $\begin{array}{rl}& \left|\frac{(1-\mu ){\displaystyle \frac{z}{f_{n,m}\left(z\right)}}+\mu {\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}}}{(1-\mu ){\displaystyle \frac{z}{f_{n,m}\left(z\right)}}+\mu }-1\right|\\ & =\mu \left|\frac{{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}}-1}{(1-\mu ){\displaystyle \frac{z}{f_{n,m}\left(z\right)}}+\mu }\right|\\ & <\frac{(1-\gamma )X\mu }{\left|(1-\mu )\exp [-(1-\gamma \left)Xw\right(z\left)\right]+\mu \right|},z\in \mathrm{U},\end{array}$ and using the fact that $|w\left(z\right)|<1$ for all $z\in \mathrm{U}$, we may easily prove that $\left|(1-\mu )\exp [-(1-\gamma \left)Xw\right(z\left)\right]+\mu \right|>\mu ,z\in \mathrm{U}.$ From the above two inequalities, it follows $\left|\frac{(1-\mu ){\displaystyle \frac{z}{f_{n,m}\left(z\right)}}+\mu {\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}}}{(1-\mu ){\displaystyle \frac{z}{f_{n,m}\left(z\right)}}+\mu }-1\right|<(1-\gamma )X,z\in \mathrm{U},$ and consequently, from (6(6) $\frac{z{f}_{\mu }^{\prime }\left(z\right)}{f_{n,m,\mu }\left(z\right)}=\frac{(1-\mu ){\displaystyle \frac{z}{f_{n,m}\left(z\right)}}+\mu {\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}}}{(1-\mu ){\displaystyle \frac{z}{f_{n,m}\left(z\right)}}+\mu }.$(6) ) we obtain $\frac{z{f}_{\mu }^{\prime }\left(z\right)}{f_{n,m,\mu }\left(z\right)}\prec 1+(1-\gamma )Xz,$ that is $f_{\mu }\in {\mathcal{S}}^{(n,m)}[X,0,\gamma ]$.

(ii) For $Y\ne 0$, we need to determine the value $r_{\ast }\in (0,1)$, such that (8) $\begin{array}{rl}& \left|\frac{(1-\mu ){\displaystyle \frac{z}{f_{n,m}\left(z\right)}}+\mu {\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}}}{(1-\mu ){\displaystyle \frac{z}{f_{n,m}\left(z\right)}}+\mu }-1\right|\\ & <\left|\left[\right(1-\gamma )X+\gamma Y]-Y\frac{(1-\mu ){\displaystyle \frac{z}{f_{n,m}\left(z\right)}}+\mu {\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}}}{(1-\mu ){\displaystyle \frac{z}{f_{n,m}\left(z\right)}}+\mu }\right|,\end{array}$(8) whenever $|z|<r_{\ast }$, which is equivalent to $\begin{array}{rl}& \left|\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}-1\right|\\ & <\left|\left(\left[\right(1-\gamma )X+\gamma Y]-Y\right)\left(\frac{1}{\mu }-1\right)\frac{z}{f_{n,m}\left(z\right)}\right.\\ & +\left.\left[\right(1-\gamma )X+\gamma Y]-Y\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\right|.\end{array}$

For $|z|<r_{\ast }$. According to the Remark 1.1 and the definition of the subordination we have that $f\in {\mathcal{S}}^{(n,m)}[X,Y,\gamma ],\mathrm{f}\mathrm{o}\mathrm{r}|z|<r_{\ast },$ is equivalent to (9) $\begin{array}{rl}\left|\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}-1\right|& <\left|\left[\right(1-\gamma )X+\gamma Y]-Y\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\right|,\\ & |z|<r_{\ast }.\end{array}$(9) Next, we will prove that (10) $\begin{array}{rl}& \left|\arg \left(\frac{z}{f_{n,m}\left(z\right)}\right)-\arg \left(\left[\right(1-\gamma )X+\gamma Y]-Y\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\right)\right|\\ & <\frac{\pi }{2},z\in \mathrm{U},\end{array}$(10) implies (11) $\begin{array}{rl}& \left|\left[\right(1-\gamma )X+\gamma Y]-Y\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\right|\\ & <\left|\left(\left[\right(1-\gamma )X+\gamma Y]-Y\right)\left(\frac{1}{\mu }-1\right)\frac{z}{f_{n,m}\left(z\right)}\right.\\ & +\left.\left[\right(1-\gamma )X+\delta Y]-Y\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\right|,z\in \mathrm{U}.\end{array}$(11) Since $f\in {\mathcal{S}}^{(n,m)}[X,Y,\gamma ]$, from Definition 1.5 and Theorem 2.1, it follows that there exist the functions $w,\tilde{w}\in \mathrm{\Omega }$, such that $\begin{array}{rl}& \left[\right(1-\gamma )X+\gamma Y]-Y\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}=\frac{(X-Y)(1-\gamma )}{1+Yw\left(z\right)}\mathrm{a}\mathrm{n}\mathrm{d}\\ & \frac{z}{f_{n,m}\left(z\right)}={\left(1+Y\tilde{w}\left(z\right)\right)}^{-\left(\right((1-\gamma )(X-Y)\left)/Y\right)}.\end{array}$ Using the above relations, the assumption (10(10) $\begin{array}{rl}& \left|\arg \left(\frac{z}{f_{n,m}\left(z\right)}\right)-\arg \left(\left[\right(1-\gamma )X+\gamma Y]-Y\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\right)\right|\\ & <\frac{\pi }{2},z\in \mathrm{U},\end{array}$(10) ) is equivalent to $\begin{array}{rl}& \left|\arg \frac{{\left(1+Y\tilde{w}\left(z\right)\right)}^{\left(-(X-Y)(1-\gamma )\right)/Y}\left(1+Yw\left(z\right)\right)}{(1-\gamma )(X-Y)}\right|<\frac{\pi }{2},\\ & z\in \mathrm{U},\end{array}$ that is $\mathrm{R}\mathrm{e}\frac{{\left(1+Y\tilde{w}\left(z\right)\right)}^{\left(-(X-Y)(1-\gamma )\right)/Y}\left(1+Yw\left(z\right)\right)}{(1-\gamma )(X-Y)}>0,z\in \mathrm{U},$ or $\begin{array}{rl}& \mathrm{R}\mathrm{e}\left[{\left(1+Y\tilde{w}\left(z\right)\right)}^{\left(-(X-Y)(1-\gamma )\right)/Y}\left(1+Yw\left(z\right)\right)\right]>0,\\ & z\in \mathrm{U}.\end{array}$ If we denote $\zeta :={\left(1+Y\tilde{w}\left(z\right)\right)}^{\left(-(X-Y)(1-\gamma )\right)/Y}\left(1+Yw\left(z\right)\right),$ the above inequality shows that $\mathrm{R}\mathrm{e}\zeta >0$, therefore $\begin{array}{rl}& |(1-\mu )\zeta +\mu |\\ & =\sqrt{{\left[\mathrm{R}\mathrm{e}(1-\mu )\zeta +\mu \right]}^2+{\left[\mathrm{I}\mathrm{m}(1-\mu )\zeta \right]}^2}>\mu ,\end{array}$ whenever $0<\mu <1$.

Dividing this inequality by $\mu >0$, we obtain $\begin{array}{rl}& \left|\left(\frac{1}{\mu }-1\right){\left(1+Y\tilde{w}\left(z\right)\right)}^{-\left(\right((1-\gamma )(X-Y)\left)/Y\right)}\right.\\ & \times \left.\left(1+Yw\left(z\right)\right)+1\right|>1,z\in \mathrm{U},\end{array}$ that is $\begin{array}{rl}& \left|\frac{(X-Y)(1-\gamma )}{1+Yw\left(z\right)}\right|\\ & <\left|\frac{(X-Y)(1-\gamma )}{1+Yw\left(z\right)}\right|\left|\left(\frac{1}{\mu }-1\right)\right.\\ & \times \left.{\left(1+Y\tilde{w}\left(z\right)\right)}^{-\left(\right((1-\gamma )(X-Y)\left)/Y\right)}\left(1+Yw\left(z\right)\right)+1\right|,\\ & z\in \mathrm{U},\end{array}$ which represents (11(11) $\begin{array}{rl}& \left|\left[\right(1-\gamma )X+\gamma Y]-Y\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\right|\\ & <\left|\left(\left[\right(1-\gamma )X+\gamma Y]-Y\right)\left(\frac{1}{\mu }-1\right)\frac{z}{f_{n,m}\left(z\right)}\right.\\ & +\left.\left[\right(1-\gamma )X+\delta Y]-Y\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\right|,z\in \mathrm{U}.\end{array}$(11) ).

From the above reasons, we will determine now the biggest value of $r_{\ast }$ such that (10(10) $\begin{array}{rl}& \left|\arg \left(\frac{z}{f_{n,m}\left(z\right)}\right)-\arg \left(\left[\right(1-\gamma )X+\gamma Y]-Y\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\right)\right|\\ & <\frac{\pi }{2},z\in \mathrm{U},\end{array}$(10) ) holds for $|z|<r_{\ast }$. Since $f\in {\mathcal{S}}^{(n,m)}[X,Y,\gamma ]$, there exists a function $w\in \mathrm{\Omega }$, such that $\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}=\frac{1+\left[\right(1-\gamma )X+\gamma Y]w\left(z\right)}{1+Yw\left(z\right)},z\in \mathrm{U},$ therefore (12) $\begin{array}{rl}& \left|\arg \left(\left[\right(1-\gamma )X+\gamma Y]-Y\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\right)\right|\\ & =\left|\arg \frac{(1-\gamma )(X-Y)}{1+Yw\left(z\right)}\right|\\ & \le \left|\arg \left[(X-Y)(1-\gamma )\right]\right|+\left|\arg \left[1+Yw\left(z\right)\right]\right|\\ & \le \arcsin \left(|Y|r\right),|z|\le r<1.\end{array}$(12)

Next, we will prove that (13) $\left|\arg \left(\frac{z}{f_{n,m}\left(z\right)}\right)\right|\le \frac{(X-Y)(1-\gamma )}{Y}\arcsin \left(Yr\right),|z|=r.$(13) Thus, from Theorem 2.1, we have $\frac{f_{n,m}\left(z\right)}{z}={\left(1+Y\tilde{w}\left(z\right)\right)}^{(X-Y)(1-\gamma )/Y},$ for some $\tilde{w}\in \mathrm{\Omega }$, and we will split this proof in the next two cases:

Case 1. If Y >0, since X − Y >0 it follows that $\begin{array}{rl}& \left|{\left(1+Y\tilde{w}\left(z\right)\right)}^{(X-Y)(1-\gamma )/y}\right|\\ & =\left|\exp \left[\frac{(X-Y)(1-\gamma )}{Y}\log \left(1+Y\tilde{w}\left(z\right)\right)\right]\right|\\ & =\exp \left[\frac{(X-Y)(1-\gamma )}{Y}\ln \left|1+Y\tilde{w}\left(z\right)\right|\right]\\ & ={\left|1+Y\tilde{w}\left(z\right)\right|}^{(X-Y)(1-\gamma )/Y}\\ & \le {\left(1+Yr\right)}^{(X-Y)(1-\gamma )/Y},|z|\le r<1.\end{array}$

Case 2. If Y <0, denoting $C:=-Y>0$, then C + X>0, and it follows that $\begin{array}{rl}& \left|{\left(1+Y\tilde{w}\left(z\right)\right)}^{(X-Y)(1-\gamma )/Y}\right|\\ & =\left|{\left[{\left(1-C\tilde{w}\left(z\right)\right)}^{-1}\right]}^{\left((1-\gamma )(X+C)\right)/C}\right|\\ & ={\left|{\left(1-C\tilde{w}\left(z\right)\right)}^{-1}\right|}^{\left((1-\gamma )(X+C)\right)/C}\\ & \le {\left(\frac{1}{1-Cr}\right)}^{\left((1-\gamma )(X+C)\right)/C}\\ & ={\left(1+Yr\right)}^{\left((1-\gamma )(X-Y)\right)/y},|z|\le r<1.\end{array}$

Combining the above two cases, we get $\begin{array}{rl}\left|\arg \left(\frac{z}{f_{n,m}\left(z\right)}\right)\right|& \le \frac{(X-Y)(1-\gamma )}{Y}\left|\arg \left(1+Yr\right)\right|\\ & \le \frac{(X-Y)(1-\gamma )}{Y}\arcsin \left(Yr\right),\\ |z|& \le r<1.\end{array}$

From (12(12) $\begin{array}{rl}& \left|\arg \left(\left[\right(1-\gamma )X+\gamma Y]-Y\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\right)\right|\\ & =\left|\arg \frac{(1-\gamma )(X-Y)}{1+Yw\left(z\right)}\right|\\ & \le \left|\arg \left[(X-Y)(1-\gamma )\right]\right|+\left|\arg \left[1+Yw\left(z\right)\right]\right|\\ & \le \arcsin \left(|Y|r\right),|z|\le r<1.\end{array}$(12) ) and (13(13) $\left|\arg \left(\frac{z}{f_{n,m}\left(z\right)}\right)\right|\le \frac{(X-Y)(1-\gamma )}{Y}\arcsin \left(Yr\right),|z|=r.$(13) ) we easily deduce that $\begin{array}{rl}& \left|\arg \left(\frac{z}{f_{n,m}\left(z\right)}\right)-\arg \left(\left[\right(1-\gamma )X+\gamma Y]-Y\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\right)\right|\\ & \le \left|\arg \left(\frac{z}{f_{n,m}\left(z\right)}\right)\right|\\ & +\left|\arg \left(\left[\right(1-\gamma )X+\gamma Y]-Y\frac{z{f}^{\prime }\left(z\right)}{f_{n,m}\left(z\right)}\right)\right|\\ & \le \arcsin \left(|Y|r\right)+\frac{(X-Y)(1-\gamma )}{Y}\arcsin \left(Yr\right)<\frac{\pi }{2},\\ & |z|\le r_{\ast },\end{array}$ where $r_{\ast }$ is given like in the assumptions (ii) and (iii) of Theorem 1.1.

The next distortion and covering theorems for the class ${\mathcal{S}}^{(n,m)}[X,Y,\gamma ]$ holds:

Theorem 2.4

If $f\in {\mathcal{S}}^{(n,m)}[X,Y,\gamma ],$ then $g\left(r\right)\le |f^{\prime }\left(z\right)|\le q\left(r\right),$ where (14) $\begin{array}{rl}g\left(r\right)& =\left\{\begin{array}{ll}\begin{array}{c}\left[1-(1-\gamma )Xr\right]\\ \exp \left[-(1-\gamma )Xr\right]\end{array},& \mathrm{i}\mathrm{f}Y=0,\\ \begin{array}{c}\frac{1-[\gamma Y+(1-\gamma \left)X\right]r}{1-Yr}\\ (1-Yr{)}^{(X-Y)(1-\gamma )/Y}\end{array},& \mathrm{i}\mathrm{f}Y\ne 0,\end{array}\right.\end{array}$(14) (15) $\begin{array}{rl}q\left(r\right)& =\left\{\begin{array}{ll}\begin{array}{c}\left[1+(1-\gamma )Xr\right]\\ \exp \left[(1-\gamma )Xr\right]\end{array},& \mathrm{i}\mathrm{f}Y=0,\\ \begin{array}{c}\frac{1+\left[\right(1-\gamma )X+\gamma Y]r}{1+Yr}\\ (1+Yr{)}^{(X-Y)(1-\gamma )/Y}\end{array},& \mathrm{i}\mathrm{f}Y\ne 0\end{array}\right.\end{array}$(15) and $|z|\le r<1$.

Proof.

Let $f\in {\mathcal{S}}^{(n,m)}[X,Y,\gamma ]$, according to Theorem 2.1 we need to distinguish the next two cases:

(i) If $Y\ne 0$, then there exists Schwarz functions, $w\in \mathrm{\Omega }$ such that $f_{n,m}\left(z\right):=z(1+Yw(z){)}^{\left((1-\gamma )(X-Y)\right)/Y}$, and by Lemma 1.2 we get (16) $\begin{array}{rl}& \frac{1-[\gamma Y+(1-\gamma \left)X\right]r}{1-Yr}{\left|Yw\left(z\right)+1\right|}^{(X-Y)(1-\gamma )/Y}\le |f^{\prime }\left(z\right)|\\ & \le \frac{1+[\gamma Y+(1-\gamma \left)X\right]r}{1+Yr}{\left|Yw\left(z\right)+1\right|}^{\left((1-\gamma )(X-Y)\right)/Y},\\ & |z|\le r<1.\end{array}$(16) Since $w\in \mathrm{\Omega }$, we have $1-|Y|r\le \left|Yw\left(z\right)+1\right|\le |Y|r+1,|z|\le r<1.$

Case 1. For Y >0. By using the fact that $-1\le Y<X\le 1$ and $0\le \gamma <1$, we have $\begin{array}{rl}& (1-|Y|r{)}^{(X-Y)(1-\gamma )/Y}\le {\left|Yw\left(z\right)+1\right|}^{(X-Y)(1-\gamma )/Y}\\ & \le (|Y|r+1{)}^{(X-Y)(1-\gamma )/Y},|z|\le r<1,\end{array}$ and from (16(16) $\begin{array}{rl}& \frac{1-[\gamma Y+(1-\gamma \left)X\right]r}{1-Yr}{\left|Yw\left(z\right)+1\right|}^{(X-Y)(1-\gamma )/Y}\le |f^{\prime }\left(z\right)|\\ & \le \frac{1+[\gamma Y+(1-\gamma \left)X\right]r}{1+Yr}{\left|Yw\left(z\right)+1\right|}^{\left((1-\gamma )(X-Y)\right)/Y},\\ & |z|\le r<1.\end{array}$(16) ) we obtain (17) $\begin{array}{rl}& \frac{1-[\gamma Y+(1-\gamma \left)X\right]r}{1-Yr}(1-|Y|r{)}^{(X-Y)(1-\gamma )/Y}\le |f^{\prime }(z)|\\ & \le \frac{1+[\gamma Y+(1-\gamma \left)X\right]r}{1+Yr}(|Y|r+1{)}^{(X-Y)(1-\gamma )/Y},\\ |z|& \le r<1.\end{array}$(17)

Case 2. If Y <0, from the fact that $-1\le Y<X\le 1$ and $0\le \gamma <1$, we have $\begin{array}{rl}& (1-|Y|r{)}^{(X-Y)(1-\gamma )/Y}\ge {\left|Yw\left(z\right)+1\right|}^{(X-Y)(1-\gamma )/Y}\\ & \ge (1+|Y|r{)}^{\left((1-\gamma )(X-Y)\right)/Y},|z|\le r<1,\end{array}$ and from (16(16) $\begin{array}{rl}& \frac{1-[\gamma Y+(1-\gamma \left)X\right]r}{1-Yr}{\left|Yw\left(z\right)+1\right|}^{(X-Y)(1-\gamma )/Y}\le |f^{\prime }\left(z\right)|\\ & \le \frac{1+[\gamma Y+(1-\gamma \left)X\right]r}{1+Yr}{\left|Yw\left(z\right)+1\right|}^{\left((1-\gamma )(X-Y)\right)/Y},\\ & |z|\le r<1.\end{array}$(16) ) we obtain (18) $\begin{array}{rl}& \frac{1-\left[\right(1-\gamma )X+\gamma Y]r}{1-Yr}(1-|Y|r{)}^{(X-Y)(1-\gamma )/Y}\ge |f^{\prime }(z)|\\ & \ge \frac{1+\left[\gamma Y\right]r+(1-\gamma )X}{1+Yr}(|Y|r+1{)}^{(X-Y)(1-\gamma )/Y},\\ & |z|\le r<1.\end{array}$(18) Now, by combining (17(17) $\begin{array}{rl}& \frac{1-[\gamma Y+(1-\gamma \left)X\right]r}{1-Yr}(1-|Y|r{)}^{(X-Y)(1-\gamma )/Y}\le |f^{\prime }(z)|\\ & \le \frac{1+[\gamma Y+(1-\gamma \left)X\right]r}{1+Yr}(|Y|r+1{)}^{(X-Y)(1-\gamma )/Y},\\ |z|& \le r<1.\end{array}$(17) ) and (18(18) $\begin{array}{rl}& \frac{1-\left[\right(1-\gamma )X+\gamma Y]r}{1-Yr}(1-|Y|r{)}^{(X-Y)(1-\gamma )/Y}\ge |f^{\prime }(z)|\\ & \ge \frac{1+\left[\gamma Y\right]r+(1-\gamma )X}{1+Yr}(|Y|r+1{)}^{(X-Y)(1-\gamma )/Y},\\ & |z|\le r<1.\end{array}$(18) ), we get (19) $\begin{array}{rl}& \frac{1-\left[\right(1-\gamma )X+\gamma Y]r}{1-Yr}(1-Yr{)}^{\left((1-\gamma )(X-Y)\right)/Y}\le |f^{\prime }(z)|\\ & \le \frac{1+[\gamma Y+(1-\gamma \left)X\right]r}{1+Yr}(1+Yr{)}^{(X-Y)(1-\gamma )/Y},\\ & |z|\le r<1.\end{array}$(19)

(ii) If Y = 0, there exists Schwarz functions, $w\in \mathrm{\Omega }$ such that $f_{n,m}\left(z\right)=z\exp \left[\right(1-\gamma \left)Xw\right(z\left)\right]$, and therefore (20) $\begin{array}{rl}& \left[1-(1-\gamma )Xr\right]\left|\exp \left[(1-\gamma )Xw\left(z\right)\right]\right|\le |f^{\prime }\left(z\right)|\\ & \le \left[(1-\gamma )Xr+1\right]\left|\exp \left[(1-\gamma )Xw\left(z\right)\right]\right|,\end{array}$(20) for $|z|\le r<1$. Since $\left|\exp \left[(1-\gamma )Xw\left(z\right)\right]\right|=\exp \left[(1-\gamma )X\mathrm{R}\mathrm{e}w\left(z\right)\right],z\in \mathrm{U}.$ By a similar way as in the previous case, we get $\begin{array}{rl}& \exp \left[-(1-\gamma )Xr\right]\le \left|\exp \left[(1-\gamma )Xw\left(z\right)\right]\right|\\ & \le \exp \left[(1-\gamma )Xr\right],|z|\le r<1.\end{array}$ Thus, (20(20) $\begin{array}{rl}& \left[1-(1-\gamma )Xr\right]\left|\exp \left[(1-\gamma )Xw\left(z\right)\right]\right|\le |f^{\prime }\left(z\right)|\\ & \le \left[(1-\gamma )Xr+1\right]\left|\exp \left[(1-\gamma )Xw\left(z\right)\right]\right|,\end{array}$(20) ) yield to $\begin{array}{rl}& \left[1-(1-\gamma )Xr\right]\exp \left[-(1-\gamma )Xr\right]\le |f^{\prime }\left(z\right)|\\ & \le \left[(1-\gamma )Xr+1\right]\exp \left[(1-\gamma )Xr\right],\end{array}$ for $|z|\le r<1$. That is complete the proof of our theorem.

Theorem 2.5

If $f\in {\mathcal{S}}^{(n,m)}[X,Y,\gamma ],$ then $|f\left(z\right)|\le \left\{\begin{array}{ll}\begin{array}{c}{\int }_0^r\left[1+(1-\gamma )X\rho \right]\\ \exp \left[(1-\gamma )X\rho \right]\mathrm{d}\rho \end{array},& \mathrm{i}\mathrm{f}Y=0,\\ \begin{array}{c}{\int }_0^r\frac{1+[\gamma Y+(1-\gamma \left)X\right]\rho }{1+Y\rho }\\ (1+Y\rho {)}^{(X-Y)(1-\gamma )/Y}\mathrm{d}\rho \end{array},& \mathrm{i}\mathrm{f}Y\ne 0,\end{array}\right.$ where $|z|\le r<1$.

Proof.

Integrated the function $f^{\prime }$ along the close segment connecting the origin with an arbitrary $z\in \mathrm{U}$, since any point of this segment is of the form $\zeta =\rho {\mathrm{e}}^{\mathrm{i}\theta }$, with $\rho \in [0,r]$, where $\theta =\arg z$ and $r=|z|$, we get $f\left(z\right)={\int }_0^z{f}^{\prime }\left(\zeta \right)\mathrm{d}\zeta ,z=r{\mathrm{e}}^{\mathrm{i}\theta },$ hence $|f\left(z\right)|=\left|{\int }_0^r{f}^{\prime }\left(\rho {\mathrm{e}}^{\mathrm{i}\theta }\right){\mathrm{e}}^{\mathrm{i}\theta }\mathrm{d}\rho \right|\le {\int }_0^r\left|f^{\prime }\left(\rho {\mathrm{e}}^{\mathrm{i}\theta }\right){\mathrm{e}}^{\mathrm{i}\theta }\right|\mathrm{d}\rho .$ Using this inequality and the right-hand side inequalities of Theorem 2.4, we need to discuss the next two cases:

1. If $Y\ne 0$, then $\begin{array}{rl}|f\left(z\right)|& \le {\int }_0^r\left|f^{\prime }\left(\rho {\mathrm{e}}^{\mathrm{i}\theta }\right){\mathrm{e}}^{\mathrm{i}\theta }\right|\mathrm{d}\rho \\ & \le {\int }_0^r\frac{1+\left[\right(1-\gamma )X+\gamma Y]\rho }{1+Y\rho }\\ & \times (1+Y\rho {)}^{\left((1-\gamma )(X-Y)\right)/Y}\mathrm{d}\rho ,\end{array}$ that is $\begin{array}{rl}|f\left(z\right)|& \le {\int }_0^r\frac{1+\left[\right(1-\gamma )X+\gamma Y]\rho }{1+Y\rho }\\ & \times (1+Y\rho {)}^{\left((1-\gamma )(X-Y)\right)/Y}\mathrm{d}\rho ,|z|\le r<1.\end{array}$

2. If Y = 0, then $\begin{array}{rl}|f\left(z\right)|& \le {\int }_0^r\left|f^{\prime }\left(\rho {\mathrm{e}}^{\mathrm{i}\theta }\right){\mathrm{e}}^{\mathrm{i}\theta }\right|\mathrm{d}\rho \\ & \le {\int }_0^r\left[1+(1-\gamma )|X|\rho \right]\exp \left[(1-\gamma )X\rho \right]\mathrm{d}\rho ,\end{array}$ that is $\begin{array}{rl}|f\left(z\right)|& \le {\int }_0^r\left[1+(1-\gamma )X\rho \right]\exp \left[(1-\gamma )X\rho \right]\mathrm{d}\rho ,\\ |z|& \le r<1.\end{array}$

Disclosure statement

No potential conflict of interest was reported by the authors.

ORCID

Fuad Alsarari  http://orcid.org/0000-0002-4759-5449

Satyanarayana Latha http://orcid.org/0000-0002-1513-8163

Teodor Bulboacă http://orcid.org/0000-0001-8026-218X